Problem: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $45.2$ years; the standard deviation is $11.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living longer than $56.4$ years.
$45.2$ $34$ $56.4$ $22.8$ $67.6$ $11.6$ $78.8$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $45.2$ years. We know the standard deviation is $11.2$ years, so one standard deviation below the mean is $34$ years and one standard deviation above the mean is $56.4$ years. Two standard deviations below the mean is $22.8$ years and two standard deviations above the mean is $67.6$ years. Three standard deviations below the mean is $11.6$ years and three standard deviations above the mean is $78.8$ years. We are interested in the probability of a bear living longer than $56.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the bears will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $34$ years and the other half $({16\%})$ will live longer than $56.4$ years. The probability of a particular bear living longer than $56.4$ years is ${16\%}$.